Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This question already has an answer here:

So I start off and assume that some $\{U_\alpha\}$ is a cover of $E$.

I want to reduce this cover to a finite subcover of $E$.

Since $p$ is a covering map it is also an open map, therefore $p(\{U_\alpha\})$ is an open cover of $B$ denote it $\{W_\alpha\}$.

But since $B$ is compact there is a finite subcover of $\{W_\alpha\}$, $\cup_{i=1}^nW_i$.

But then since $p$ is continuous $p^{-1}(\cup_{i=1}^nW_i)$ is an open cover of $E$ which must stay finite since $p^{-1}(b)$ is finite for all $b$.

share|cite|improve this question

marked as duplicate by Najib Idrissi, Sami Ben Romdhane, user127.0.0.1, amWhy, Davide Giraudo Apr 16 '14 at 15:58

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

duplicate of – Henno Brandsma Apr 16 '14 at 5:27

2 Answers 2

You haven't used that $B$ is compact. Hint: project your open cover of $E$ to an open cover of $B$...

share|cite|improve this answer
If we project our open cover of $E$ to $B$ we get a cover of $B$ and this can be reduced to a finite subcover since $B$ is compact but I don't see how to then send this back over to find a finite subcover of $E$. – EgoKilla Apr 15 '14 at 18:12
@EgoKilla This is where you use your finite inverse image condition. – Mike Miller Apr 15 '14 at 18:13
Is my edited question a proper solution now? – EgoKilla Apr 15 '14 at 18:26
@EgoKilla You need to be a bit more careful. You showed that $E$ has a finite open cover; but you need to show that your original cover has a finite subcover. You're close, though. – Mike Miller Apr 15 '14 at 18:39
As others have suggested, letting $p\{U_{\alpha}\}$ be your cover for $B$ won't quite work. The problem is, you want to create an open cover for $B$ consisting of open sets whose inverse images are 'well-behaved' (more explicitly, you want $p^{-1}(W)$ to be covered by finitely many $U_{\alpha}$). – Alex Zorn Apr 16 '14 at 14:56

You could use the fact that a covering map with finitefibers is a perfect map, i.e. a surjective and closed map with compact fibers. Closedness of $p$ is shown as follows:
Let $C\subset E$ be closed, $b\notin p[C]$. Then $p^{-1}(b)=\{b_1,...,b_n\}$ is a finite subset of $E-C$. There is an evenly covered neighborhood $U$ of $b$ whose preimage is the union of $U_i$, $i=1,....,n$ so that $b_i\in U_i$. Now each $b_i$ has a neighborhood $V_i$ contained in $U_i\cap(E-C)$. If $x\in \bigcap_i p[V_i]$, then its fiber has one point in each $U_i$ which must thus be in $V_i$, so $x$ cannot be in $p[C]$.

Now perfect maps are proper, meaning that preimages of compact sets are compact.

Also see my answer for the more general case of a fiber bundle with compact fiber $F$. A covering space can be characterized as a fiber bundle where the fiber is discrete.

Instead of looking at the images of the $U_\alpha$, consider the following sets $$U^b=\text{ a finite subcover of $p^{-1}(b)$ from the $U_\alpha$}\\ W_b=p_*(U^b)=B-f(E-U^b)=\{c\in B\mid f^{-1}(c)\subseteq U^b\}$$ This $p_*$ is also called the dual image and it is inclusion preserving. It's easy to see that $p^{-1}(W_b)\subseteq U^b$. Furthermore, since $p$ is closed, then $p_*(U^b)$ is an open neighborhood of $b$, so the $W_b,\ b\in B$, form an open cover of $B$. There are now $b_1,...,b_n$, so that $W_{b_i}$ cover $B$. It follows that the $U^{b_i}$ cover $E$.

share|cite|improve this answer
Instead of this solution can you provide me with some more hints on how to modify my above solution? – EgoKilla Apr 16 '14 at 10:16
Can you please help me modify my solution? I would like to finish the proof the way I began it. Once I reduce to a finite subcover of $B$ what kind of trick can I use to send it back across as a finite subcover of the $\{U_\alpha\}$? – EgoKilla Apr 16 '14 at 13:55
@EgoKilla: I don't see a way of doing this without using the closedness of $p$. Note that my added proof resembles the proof that perfect maps are proper. – Stefan Hamcke Apr 16 '14 at 14:35

Not the answer you're looking for? Browse other questions tagged or ask your own question.